Question:

<p><strong>Step 1:</strong> Find the intersection points of line $x + y = 1$ and circle $x^2 + y^2 = 9$.</p><p>From the line: $y = 1 - x$</p><p>Substitute into circle: $x^2 + (1-x)^2 = 9$</p><p>$x^2 + 1 - 2x + x^2 = 9$</p><p>$2x^2 - 2x - 8 = 0$</p><p>$x^2 - x - 4 = 0$</p><p>The intersection points exist (discriminant = $1 + 16 = 17 > 0$).</p><p><strong>Step 2:</strong> Use the family of circles method. Any circle through the intersection points of a line $L$ and circle $C$ can be written as:</p><p>$C + \lambda L = 0$</p><p>$(x^2 + y^2 - 9) + \lambda(x + y - 1) = 0$</p><p>$x^2 + y^2 + \lambda x + \lambda y - 9 - \lambda = 0$</p><p><strong>Step 3:</strong> For the smallest circle, the diameter equals the chord length. The smallest circle has its center on the perpendicular bisector of the chord and passes through both intersection points.</p><p>The perpendicular to $x + y = 1$ has slope $1$. The center lies on line $y - y_0 = 1(x - x_0)$, or equivalently, the center must satisfy that the circle is orthogonal to a degenerate configuration.</p><p><strong>Step 4:</strong> For the smallest circle through two points on a line, the center must be equidistant from both points with minimum distance. This occurs when we minimize the radius.</p><p>From $x^2 + y^2 + \lambda x + \lambda y - 9 - \lambda = 0$:</p><p>Center: $\left(-\frac{\lambda}{2}, -\frac{\lambda}{2}\right)$</p><p>Radius$^2 = \frac{\lambda^2}{4} + \frac{\lambda^2}{4} + 9 + \lambda = \frac{\lambda^2}{2} + 9 + \lambda$</p><p><strong>Step 5:</strong> Minimize radius$^2$ by taking derivative with respect to $\lambda$:</p><p>$\frac{d}{d\lambda}\left(\frac{\lambda^2}{2} + \lambda + 9\right) = \lambda + 1 = 0$</p><p>$\lambda = -1$</p><p><strong>Step 6:</strong> Substitute $\lambda = -1$ into the family equation:</p><p>$x^2 + y^2 - x - y - 9 + 1 = 0$</p><p>$x^2 + y^2 - x - y - 8 = 0$</p><p><strong>∴ Answer: B</strong></p>